foass77W
2021-01-27
Answered

Let G be a group of order $p}^{m$ where p is prime number and m is a positive integer. Show that G contains an element of order p.

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oppturf

Answered 2021-01-28
Author has **94** answers

Let G be a group of order

Then,

Where

Then, G is a finite group and the prime p divides the order of G, i.e.,

Proving the result by induction on n,

Let n=1

Then,

Then, |G|=1

Thus, the group has only one element of order 1. The result holds for n=1

Next, suppose every group with pk elements contains an element of order p for every k Consider the class equation of G,

If G=Z(G), then G is abelian and hence the result follows.

Suppose

Then, for every a !in Z(G),

If for some

Otherwise, for every

Then,

Thus, it follows that Z(G) and hence G contains an element of order p.

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We're trying to understand the field $\frac{{\mathbb{Z}}_{2}\left[x\right]}{\u27e8{x}^{3}+{x}^{2}+1\u27e9}$ , and in particular why it has eight elements. I know that ${x}^{3}+{x}^{2}+1$ is irreducible in $\mathbb{Z}}_{2$ and so factor is going to be a field.

The issue I'm having is that the notes claim that$(x+I)$ has an inverse in the field, where $I=\u27e8{x}^{3}+{x}^{2}+1\u27e9$ . They write

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I don't understand the last line, I know that multiplication of cosets is defined that way, I just don't see why we get 1 at the end. I think it's possible I didn't fully understand cosets and ideals but I can't fully articulate what the issue is. Any help towards filling in the gaps of my understanding would be greatly appreciated.

The issue I'm having is that the notes claim that

I don't understand the last line, I know that multiplication of cosets is defined that way, I just don't see why we get 1 at the end. I think it's possible I didn't fully understand cosets and ideals but I can't fully articulate what the issue is. Any help towards filling in the gaps of my understanding would be greatly appreciated.

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